Answer:
Expression: N = C·L·l(t)· T + 20
The initial value problem and solution are expressed as a first order differential equation.
Step-by-step explanation:
First, gather the information:
total population, N = 2 000
Proportionality constant, C = 0.0002
l(t) number of infected individuals = l(t)
healthy individuals = L
The equation is given as follows:
N = C·L·l(t)
However, there is a change with time, so the expression will be:
[tex]\frac{dN}{dt}[/tex] = C·L·l(t)
multiplying both sides by dt gives:
dN = C·L·l(t)
Integrating both sides gives:
[tex]\int\limits^a_b {dN} \, dt[/tex] = [tex]\int\limits^a_b {CLl(t)} \, dt[/tex]
N = C·L·l(t)· T + K
initial conditions:
T= 0, N₀ = (0.01 ₓ 2 000) = 20
to find K, plug in the values:
N₀ = K
20 = K
At any time T, the expression will be:
N = C·L·l(t)· T + 20 Ans
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(t)=44t−0.83t2h(t)=44t-0.83t2. Find the average velocity over the given time intervals.
a. [3, 4]:
b. [3, 3.5]:
c. [3, 3.1]:
d. [3, 3.01]:
e. [3, 3.001]
Answer:
a. 38.19m/s
b. 38.605m/s
c. 38.937m/s
d. 39.0117m/s
e. 39.01917m/s
Step-by-step explanation:
The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:
[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]
Where:
[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]
a. Let's find h(3) and h(4) using the data provided by the problem:
[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]
The average velocity over the interval [3, 4] is :
[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]
b. Let's find h(3.5) using the data provided by the problem:
[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]
The average velocity over the interval [3, 3.5] is :
[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]
c. Let's find h(3.1) using the data provided by the problem:
[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]
The average velocity over the interval [3, 3.1] is :
[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]
d. Let's find h(3.01) using the data provided by the problem:
[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]
The average velocity over the interval [3, 3.01] is :
[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]
e. Let's find h(3.001) using the data provided by the problem:
[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]
[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]
The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.
Given :
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].
a) [3,4]
At time t = 3 and 4, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]
b) [3,3.5]
At time t = 3 and 3.5, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]
c) [3,3.1]
At time t = 3 and 3.1, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]
d) [3,3.01]
At time t = 3 and 3.01, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]
e) [3,3.001]
At time t = 3 and 3.001, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]
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Find the direction cosines and direction angles of the given vector. (Round the direction angles to two decimal places.) a = 5, 9, 3 cos(α) = cos(β) = cos(γ) = α = ° β = ° γ = °
Answer:
Step-by-step explanation:
given is a vector as (5,9,3)
a = (5,9,3)
To find out direction cosines
First let us calculate modulus of vector a
[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]
Direction ratios are (5,9,3)
Magnitude of vector a = [tex]\sqrt{115}[/tex]
So direction cosines would be
[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
Angles would be
[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
=cos inverse (0.4662, 0.8393, 0.2798)
= (62.21, 32.93,32,94)
Two students, X and Y, forgot to put their names on their exam papers. The professor knows that these two students do well on the exam with probabilities 0.8 and 0.4, respectively. After grading, the professor notices that X and Y forgot to put their names on their exams. One of their exams was done well and the other was done poorly. Given this information, and assuming that students worked independently of each other, what is the probability that the good exam belongs to student X
Answer:
The probability that the good exam belongs to student X is 0.8571.
Step-by-step explanation:
It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,
Compute the probability that exactly one student does well in the exam as follows:
[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]
Then the probability that X is the one who did well in the exam is:
[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]
Thus, the probability that the good exam belongs to student X is 0.8571.
A dead body was found within a closed room of a house where the temperature was a constant 65° F. At the time of discovery the core temperature of the body was determined to be 85° F. One hour later a second measurement showed that the core temperature of the body was 80° F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found.
Answer:
1 hr 52 minutes
Step-by-step explanation:
As per Newton law of cooling we have
[tex]T(t) = T_s +(T_0-T_s)e^{-kt}[/tex]
where T0 is the initial temperature of the body
Ts = temperature of surrounding
t = time lapsed
k = constant
Using this we find that T0 = 98.6 : Ts= 65
Let x hours be lapsed before the body was found.
Then we have
[tex]T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917[/tex]
Next after 1 hour temperature was 80
[tex]T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464[/tex]
Dividing we get
[tex]e^k = 1.325408\\k = 0.2817[/tex]
Substitute this in
[tex]e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863[/tex]
approximately 1 hour 52 minutes have lapsed.
Exercise 1.28. We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color? (a) Assume that the balls are sampled without replacement. (b) Assume that the balls are sampled with replacement. (c) When is the answer to part (b) larger than the answer to part (a)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?
Answer:
Step-by-step explanation:
given that we have an urn with m green balls and n yellow balls. Two balls are drawn at random.
a) Assume that the balls are sampled without replacement.
m green and n yellow balls
For 2 balls to be drawn at the same colour
no of ways = either 2 green or 2 blue = mC2+nC2
Total no of ways = (m+n)C2
Prob =
= [tex]\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}[/tex]
=[tex]\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}[/tex]
B) Assume that the balls are sampled with replacement
In this case, probability for any draw for yellow or green will be constant as
n/M+n or m/m+n respectively
Reqd prob = [tex](\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2[/tex]
=[tex]\frac{m^2+n^2}{(m+n)^2}[/tex]
c) Part B prob will be more than part a because with replacement prob is more than without replacement.
II time drawing same colour changes to m-1/.(m+n-1) if with replacement but same as m/(m+n) without replacement
[tex]\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0[/tex]
Since n>0 is true always, b is greater than a.
The question explores the concept of probability within scenarios of drawing balls of different colors from an urn, with and without replacement. It explores how the number of balls left in the urn changes the likelihood of drawing two balls of the same color. The answer is calculated using mathematical odds and conditions, showing that replacement affects probability especially when the total number of items (balls in this case) is small.
Explanation:The subject of this question is probability, specifically conditional probability and probability with and without replacement. Here are the calculations needed to answer the question:
(a) When the balls are drawn without replacement, the probability that the two balls drawn have the same color is the sum of the probability of drawing two green balls and the probability of drawing two yellow balls. The probability of drawing two green balls is (m/(m+n)) * ((m-1)/(m+n-1)). Similarly, the probability of drawing two yellow balls is (n/(m+n)) * ((n-1)/(m+n-1)). The sum of these two probabilities gives the required probability. (b) When the balls are drawn with replacement, the same logic applies; however, since the balls are replaced, the denominator term doesn't decrease for the second draw. Thus, the probability of drawing two green balls is (m/(m+n)) * (m/(m+n)), and the probability of two yellow balls is (n/(m+n)) * (n/(m+n)). (c) The answer to part (b) becomes larger than the answer to part (a) when m and n are small numbers. This is because, when m and n are small, the probability of drawing a similarly colored ball in the second draw becomes more significant if the ball is replaced after the first draw, compared to if it is not replaced, causing the probability with replacement to be higher.
In a nutshell, the calculation for probability tells us how likely an outcome is, given the mathematical odds and conditions (in this case, if the balls are replaced or not after drawing).
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The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a sample of size 100, resulting in the following frequencies:
Number of particles: 0, 1, 2, 3, 4, , 5, 6, 7
Frequency: 1, 2, 3, 12, 11, 15, 18, 10
Number of particles: 8, 9, 10, 11, 12, 13, 14
Frequency: 12, 4, 5, 3, 1, 2, 1
(a.) What proportion of the sampled wafers had at least one particle? At least five particles?
(b.) What proportion of the sampled wafers had between five and ten particles, inclusive? Strictly between five and ten particles?
(c.) Draw a histogram using relative frequency on the vertical axis. How would you describe the shape of the histogram.
The proportion of sample wafers that had at least one particle is 96%, those with at least five particles is 71%. For wafers between five and ten particles (inclusive), the proportion is 64%, and strictly between five and ten particles it is 44%. The histogram based on the given data would appear roughly bell-shaped with a right skew.
Explanation:(a.) To calculate the proportion of sampled wafers that had at least one particle, we sum the frequencies of all groups with one or more particles and divide by the total sample size. So, the proportion is given by: (2+3+12+11+15+18+10+12+4+5+3+1+2+1)/100 = 96/100 = 0.96 or 96%.
Similarly, for wafers with at least five particles, the calculation is: (15+18+10+12+4+5+3+1+2+1)/100 = 71/100 = 0.71 or 71%.
(b.) For wafers with between five and ten particles (inclusive), we sum frequencies from 5 to 10 particles, giving (15+18+10+12+4+5)/100 = 64/100 = 0.64 or 64%.
For wafers strictly between five and ten particles (i.e., 6 to 9 particles inclusive), the calculation is: (18+10+12+4)/100 = 44/100 = 0.44 or 44%.
(c.) A histogram with relative frequency would show the proportion of wafers (y-axis) against the number of particles (x-axis). Given the distribution of frequencies, it might be expected that the histogram appears roughly bell-shaped, though the peak would be skewed to the right considering higher numbers around 5 to 8 particles.
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According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.
Answer:
The mean number of adults who would have bank savings accounts is 32.
Step-by-step explanation:
For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
In this problem, we have that:
[tex]p = 0.64[/tex]
If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.
This is E(X) when [tex]n = 50[/tex].
So
[tex]E(X) = np = 50*0.64 = 32[/tex]
The mean number of adults who would have bank savings accounts is 32.
You would like to make a nutritious meal of eggs, mixed vegetables and brown rice. The meal should provide at least 35 g of carbohydrates, at least 30 g of protein, and no more than 45 g of fat. One serving of eggs contains 2 g of carbohydrates, 18 g of protein, and 12 g of fat. A serving of mixed vegetables contains 14 g of carbohydrates, 15 g of protein, and 8 g of fat. A serving of rice contains 40 g of carbohydrates, 6 g of protein, and 1 g of fat. A serving of eggs costs $3.75, a serving of mixed vegetables costs $3.50, and a serving of rice costs $2. It is possible to order a partial serving, e.g. 0.75 servings of rice. Formulate a linear optimization model that could be used to determine the number of servings of eggs, mixed vegetables, and rice for your meal that meets the nutrition requirements at minimal cost.
Answer:
Let G be the number of eggs in the meal
V be the number of servings of mixed vegetables in the meal
R be the number of servings of brown rice in the meal
Objective function = Minimize 3.75G + 3.50V + 2R
Constraints:
2G + 14V + 40R ≥ 35(Carbohydrates)
18G + 15V + 6R ≥ 30(Protein)
12G + 8V + R ≤ 45(Fat)
G, M, B ≥ 0
Indicate in standard form the equation of the line passing through the given points.
L(5.0), M(0,5)
Answer:
y = - x + 5
Step-by-step explanation:
L(5.0), M(0,5)
y = mx + b
m = (5 - 0) / (0 - 5) = 5 / -5 = - 1
b = y - mx = 5 - ((-1) x 0) = 5
y = - x + 5
How many observations should be made if she wants to be 86.64 percent confident that the maximum error in the observed time is .5 second? Assume that the standard deviation of the task time is four seconds.
Answer:
144 observations
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.8684}{2} = 0.0668[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.0668 = 0.9332[/tex], so [tex]z = 1.5[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
In this problem, we have that:
[tex]\sigma = 4, M = 0.5[/tex]
We want to find n
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.5*\frac{4}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 6[/tex]
[tex]\sqrt{n} = \frac{6}{0.5}[/tex]
[tex]\sqrt{n} = 12[/tex]
[tex]\sqrt{n}^{2} = (12)^{2}[/tex]
[tex]n = 144[/tex]
A sports statistician is interested in determining if there is a relationship between the number of home team and visiting team losses and different sports. A random sample of 526 games is selected and the results are given below. Find the critical value chi Subscript alpha Superscript 2 to test the claim that the number of home team and visiting team losses is independent of the sport. Use alphaequals0.01. Round to three decimal places.
Answer:
The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"
[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.29,3,TRUE)"
Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
F B S Bs Total
home wins 39 156 25 83 303
Visitor wins 31 98 19 75 223
Total 70 254 44 158 526
We need to conduct a chi square test in order to check the following hypothesis:
H0: The number of home team and visiting team losses is independent of the sport.
H1: The number of home team and visiting team losses is dependent of the sport.
The level of significance assumed for this case is [tex]\alpha=0.01[/tex]
The statistic to check the hypothesis is given by:
[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
And the calculations are given by:
[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]
[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]
[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]
[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]
[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]
[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]
[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]
[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]
And the expected values are given by:
F B S Bs Total
home wins 40.32 146.32 25.35 91.02 303
Visitor wins 29.68 107.68 18.65 66.98 223
Total 70 254 44 158 526
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]
The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.29,3,TRUE)"
Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.
Use the concept thaty = c, −[infinity] < x < [infinity],is a constant function if and only ify' = 0to determine whether the given differential equation possesses constant solutions.9xy' + 5y = 10
Answer:
Yes, one of the solutions of this differential equation is a constant solution equation.
Step-by-step explanation:
9xy' + 5y = 10.
If y' = 0, 9xy' = 0
5y = 10
y = 2 = c.
So, for all real values of x such that -∞ < x < ∞, 9xy' will be 0 and one of the solutions of the differential equation will be y = 2.
Hope this helps!
According to a recent report, 60% of U.S. college graduates cannot find a full time job in their chosen profession. Assume 57% of the college graduates who cannot find a job are female and that 18% of the college graduates who can find a job are female. Given a male college graduate, find the probability he can find a full time job in his chosen profession? (See exercise 58 on page 220 of your textbook for a similar problem.)
Answer:
There is a 55.97% that a male can find a full time job in his chosen profession.
Step-by-step explanation:
We have these following probabilities:
A 60% probability that a college graduates cannot find a full time job in their chosen profession.
A 40% probability that a college graduates can find a full time job in their chosen profession.
57% of the college graduates who cannot find a job are female
43% of the college graduates who cannot find a job are male
18% of the college graduates who can find a job are female
82% of the college who can find a job are male.
Given a male college graduate, find the probability he can find a full time job in his chosen profession?
The total males are 43% of 60%(Those who cannot find a job) and 82% of 40%(Those who can find a job). So the percentage of males is [tex]P(M) = 0.43*0.60 + 0.82*0.40 = 0.586[/tex]
Those who are males and find a job in their chosen profession are 82% of 40%. So [tex]P(M \cap J) = 0.82*0.40 = 0.328[/tex]
[tex]P = \frac{P(M \cap J)}{P(M)} = \frac{0.328}{0.586} = 0.5597[/tex]
There is a 55.97% that a male can find a full time job in his chosen profession.
In a study of environmental lead exposure and IQ, the data was collected from 148 children in Boston, Massachusetts. Their IQ scores at age of 10 approximately follow a normal distribution with mean of 115.9 and standard deviation of 14.2. Suppose one child had an IQ of 74. The researchers would like to know whether an IQ of 74 is an outlier or not.
Calculate the lower fence for the IQ data, which is the lower limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the lower fence.
Answer:
a) Lower inner fence = 77.6168 = 77.62 to 2 d.p
Lower outer fence = 48.9044 = 48.90 to 2 d.p
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159
Step-by-step explanation:
Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Q₁ = 25th percentile = lower quartile
IQR = Inter quartile Range = Q₃ - Q₁
Q₃ = 75th percentile = upper quartile
To calculate Q₁ for a normal distribution with only mean and standard deviation known,
We need the standardized score whose probability is 0.25 P(z) = 0.25
From the normal distribution table
z = (± 0.674)
z = (x - xbar)/σ
x = the value in the data we're interested in,
xbar = mean = 115.9
σ = standard deviation = 14.2
Lower quartile corresponds to (z = - 0.674)
- 0.674 = (x - 115.9)/14.2
Q₁ = X = 106.3292
The upper quartile, Q₃ corresponds to z = (+0.674)
Q₃ = 125.4708
IQR = 125.4708 - 106.3292 = 19.1416
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168
Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)
We standardize 74 by obtaining its z-score
z = (x - xbar)/σ
z = (74 - 115.9)/14.2 = - 2.95
P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)
Final answer:
The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.
Explanation:
To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.
The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:
Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5
Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.
The standard deviation, the range, and the interquartile range (IQR) summarize the variability of the data. a. Why is the standard deviation usually preferred over the ranges? b. Why is the IQR sometimes preferred to the standard deviation? c. What is an advantage of the standard deviation over the IQR?
Answer:
Step-by-step explanation:
a) The standard deviation is usually preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when they are outliers,and the standard deviation uses all of the data.
b)The IQR sometimes referred to the standard deviation when there is an outlier because the IQR is less sensitive to this features than standard deviation.
that is the IQR is not affected by an outlier,while the standard deviation is affected by an outlier.
c)The advantage of the standard deviation over the IQR is the standard deviation takes into account the values of all observation,while the IQR uses only some of the data.
An experiment was performed upon rats to investigate the effect of ingesting Alar (a chemical sprayed on apple trees to keep fruit from dropping before ripe) upon subsequent cancer rates.
The following variables were measured:
gender (0=female, 1=male); weight (g); dose of Alar (nil, low, high); and number of tumors.
Which of the following is FALSE?
A) Gender is categorical; dose is ordinal
B) Gender is discrete; weight is continuous
C) Number of tumors is categorical
D) Dose is discrete
E) Weight is continuous
Answer:
Option C and D are false
Step-by-step explanation:
All the mentioned option are correct in the given scenario except option C and D.
The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.
Now consider all options
A) Gender is categorical ; dose is ordinal
This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.
B) Gender is discrete; weight is continuous
This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.
Option C and D are already discussed an option E is discussed in option B.
The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.
Explanation:In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.
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Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?
Answer:
[tex] P(X>9) = 0.3593[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=25, p=0.3089)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
For this case we want this probability:
[tex] P(X >9)[/tex]
And we can use the complement rule like this:
[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:
[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]
[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]
[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]
[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]
[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]
[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]
[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]
[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]
[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]
And in order to do the operations we can use the following excel code:
"=1-BINOM.DIST(8,25,0.3089,TRUE)"
And we got:
[tex] P(X>9) = 0.3593[/tex]
Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes. What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?
Answer: 1
Step-by-step explanation:
If a random variable x is uniformly distributed in [a,b] the
Mean = [tex]\dfrac{a+b}{2}[/tex]
Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]
Let x = Times required for a cable company to fix cable problems
As per given.
x is uniformly distributed between 40 minutes and 65 minutes.
Then , mean = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes
Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]minutes
Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )
= P(52.5-2(7.22)< X < 52.5+2(7.22))
=P(38.06 <X < 66.94 ).
But x lies between 40 minutes and 65 minutes.
Also, [40 minutes, 65 minutes]⊂ [38.06 minutes , 66.94 minutes]
Therefore ,P(38.06 <X < 66.94 ) =1
∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.
The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,
Z scoreZ score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
z = (x - μ)/σ
where x is the raw score, μ is the mean and σ is the standard deviation.
The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.
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The perimeter of the window of the camper shell is 130 in. Find the length of one of the shorter sides of the window.
in
You can't deduce the length of a side from the perimeters of a rectangle.
Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.
So, we know that
[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]
But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.
The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.
Explanation:To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.
However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.
Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.
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Solve, graph, and give interval notation for the inequality:
4(3x − 4) < 32 AND 2x + 1 ≤ 8x + 25
Answer:
The answer to your question is
Step-by-step explanation:
Inequality 1
4(3x - 4) < 32
12x - 16 < 32
12x < 32 + 16
12x < 48
x < 48/12
x < 4
Inequality 2
2x + 1 ≤ 8x + 25
2x - 8x ≤ 25 - 1
- 6x ≤ 24
x ≥ 24/-6
x ≥ - 4
- See the graph below
- Interval [-4, 4)
To solve the inequality system, we divide both sides of each inequality by the respective coefficient to isolate the variable. The solutions are x < 4 and -4 ≤ x. The graph of the solution is a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.
Explanation:To solve the inequality 4(3x - 4) < 32, we divide both sides of the inequality by 4 to isolate the variable. This gives us 3x - 4 < 8. Adding 4 to both sides of the inequality gives us 3x < 12.
Finally, dividing both sides of the inequality by 3 gives us x < 4.
For the inequality 2x + 1 ≤ 8x + 25, we subtract 2x from both sides of the inequality to isolate the variable. This gives us 1 ≤ 6x + 25. Subtracting 25 from both sides of the inequality gives us -24 ≤ 6x.
Finally, dividing both sides of the inequality by 6 gives us -4 ≤ x.
The solutions to the inequality system are x < 4 and -4 ≤ x. The graph of the solution would be a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.
The interval notation for the solution is (-∞, 4) and [-4, ∞).
Please help, thank you!
Two numbers total 63 and have a difference of 11. Find the two numbers.
Answer:
37 and 26
Step-by-step explanation:
Let the bigger number be x and the smaller number be y.
Then the two numbers total 63 will give us:
y+x=63------>eqn1
The two numbers have a difference of 11.
y-x=11------->eqn2
We add equations 1 and 2 to obtain:
2y=74
Divide both sides to get:
y=37
Put y=37 into eqn2 to get:
37-x=11
x=37-11
x=26
This larger number is 37 and the smaller number is 26
A woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard. The injury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in $1000s) 36, 62, 73, 114, 139, 140, 148, 154, 238, 290, 340, 410, 600, 750, 750, 750,1050, 1100, 1135, 1150, 1200, 1200, 1250, 1578, 1700, 1825, and 2000, from which?xi = 20,182, ?xi2 = 24,656,384.What is the maximum possible amount that could be awarded under the two-standard-deviation rule? (Round your answer to the nearest whole number.)
Answer:
variance = (27*24656384-20182^2)/(27*26) =368104.3
standard devaition SD= sqrt(368104.3) =606.716
maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD
= (20182/27)+(2*606.716)
= 1960.913
=$1960913
Express the negations of each of these statements so that all negation symbols immediately precede predicates. a. ∃z∀y∀xT (x, y, z) b. ∃x∃yP (x, y) ∧ ∀x∀yQ(x, y) c. ∃x∃y(Q(x, y) ↔ Q(y, x)) d. ∀y∃x∃z(T (x, y, z) ∨ Q(x, y))
Final answer:
The negations of the given predicate logic statements are expressed by switching the quantifiers (∀ and ∃) and adding negation symbols (¬) directly before the predicates, resulting in new statements that oppose the original ones.
Explanation:
The goal is to express the negation of each of the given predicate logic statements such that all negation symbols immediately precede predicates.
For the statement ∃z∀y∀xT(x, y, z), the negation would be ∀z∃y∃x¬T(x, y, z), which states that there is no z for which every y and every x make T(x, y, z) true.The negation of ∃x∃yP(x, y) ∧ ∀x∀yQ(x, y) is ∀x∀y¬P(x, y) ∨ ∃x∃y¬Q(x, y), meaning there are no such x and y that P(x, y) is true or there exists some x and y for which Q(x, y) is not true.For ∃x∃y(Q(x, y) ↔ Q(y, x)), the negation would be ∀x∀y¬(Q(x, y) ↔ Q(y, x)), indicating that for all x and y, it is not the case that Q(x, y) if and only if Q(y, x).The negation of the statement ∀y∃x∃z(T(x, y, z) ∨ Q(x, y)) is ∃y∀x∀z(¬T(x, y, z) ∧ ¬Q(x, y)), stating there exists a y such that for all x and z, neither T(x, y, z) nor Q(x, y) are true.100pts: What is the remainder when 3^128 is divided by 17?
Answer:
The remainder is 9
Step-by-step explanation:
3^128 is divided by 17
find the value of 3^128 first.
3^128 = 1.17901845777E61
Then you divide by 17
1.17901845777E61 ÷ 17
= 6.93540269279E59
Approximately 6. 9340
6 remainder 9
Answer:
6 remander 1
Step-by-step explanation:
first you solve 3^128 which equals E61 then divide it by 17 which equals E59
A sociologist is studying the effect of having children within the first two years of marriage on the divorce rate. Using hospital birth records, she selects a random sample of 200 couples that had a child within the first two years of marriage. Following up on these couples, she finds that 80 couples are divorced within five years. Use Scenario 8-4. A 90% confidence interval for the proportion p of all couples that had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056.a. All the answers are correct.b. Based on this interval, we can clearly see that the divorce rate is well below the 50% national average for all marriages.c. At the 10% alpha level, we would reject the claim that the divorce rate is 50% for couples who had a child within the first two years of marriage.d. Based on this interval, we can clearly see that the divorce rate is between 35% and 46%.
Answer:
The correct option is (a).
Step-by-step explanation:
The hypothesis of the study can be defined as:
H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50
Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50
The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)
The confidence level is 90%, the significance level (α) is:
[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]
Decision Rule:
If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.
Interpretation of the Confidence interval:
The confidence interval is (35%, 46%), this implies divorce rate is less than 50% for couples who had a child within the first two years of marriage.At 10% significance level, the null hypothesis will be rejected stating that the divorce rate is different from 50% for couples who had a child within the first two years of marriage.The confidence interval clearly interprets that 90% of the divorce rate for couples who had a child within the first two years of marriage is between 35% and 46%.Thus all the options are correct.
The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.
Explanation:Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.
This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.
Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.
Defects in a product occur at random according to a Poisson distribution with parameter ???? = 0.04. What is the probability that a product has one or more defects? If the manufacturing process of the product is improved and the occurrence rate of defects is cut in half to ???? = 0.02. What effect does this have on the probability that the product has one or more defects?
Answer:
When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically (0.0392 to 0.0198), to almost the half of its original value too.
Step-by-step explanation:
Poisson distribution formula
P(X=x) = f(x) = (λˣe^(-λ))/x!
λ = 0.04.
And the probability that a products one or more defects is the same thing as 1 minus the probability that a product has no defect.
P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)
P(X ≥ 1) = 1 - (0.04⁰e^(-0.04))/0! = 1 - 0.9608 = 0.0392
When the occurrence rate of defect is cut in half, that is, λ = 0.02,
P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)
P(X ≥ 1) = 1 - (0.02⁰e^(-0.02))/0! = 1 - 0.9802 = 0.0198
When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically, to almost the half of its original value too.
Hope this helps!
A 25-foot ladder is leaning against a house with the base of the ladder 5 feet from the house. How high up the house does the ladder reach? Round to the nearest tenth of a foot. The ladder reaches feet up the side of the house
Answer: 24 feet
Step-by-step explanation:
By using Pythagoras rule:
Let x be the high up the house does the ladder reached.
X^2 + 5^2= 25^2
X^2 = 25^2 - 5^2
x^2 = 625 - 25
x^2 = 600
Square both side
x = sqrt(600)
x= 24.495
x = 24 feet
Find the distance between the points (-5, -10) and (2, 4).
Math item stem image
CLEAR CHECK
4.58
12.12
15.65
21
Answer:
15.65
Step-by-step explanation:
Suppose we have two points:
[tex]A = (x_{1}, y_{1})[/tex]
[tex]B = (x_{2}, y_{2})[/tex]
The distance between these points is:
[tex]D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]
So, for points (-5, -10) and (2, 4)
[tex]D = \sqrt{(2 - (-5))^{2} + (4 - (-10))^{2}}[/tex]
[tex]D = \sqrt{7^{2} + 14^{2}[/tex]
[tex]D = 15.65[/tex]
So the correct answer is:
15.65
As a result of discharges from local dry cleaner, dinitrotoluene concentration in the groundwater is 8 mg/L. RfD for dinitrotoluene is 2.0 x 10-3 mg/kg-day. The average 70 Kg person drinks 2L/day water. The hazard ratio is most nearly:
Answer:
114.3
Step-by-step explanation:
If a 70kg person ingests 2L of water per day containing 8 mg/L of dinitrotoluene, the concentration of dinitrotoluene on that person's body is:
[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]
The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)
[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]
The hazard ratio is most nearly 114.3.
In an experimental study on friendliness and tipping, every alternate customer to whom the waiter is extra friendly toward are referred to as the...A. Control group
B. Experimental group
C. Nonexperimental group
D. Dependent group
Answer:
A. Control group
Step-by-step explanation:
A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.
An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.
Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.